cylinder with a moment of inertia I (about its axis of symmetry), mass m, and ra
ID: 2238531 • Letter: C
Question
cylinder with a moment of inertia I (about its axis of symmetry), mass m, and radius r has a massless string wrapped around it which is tied to the ceiling . At time t=0 the cylinder is released from rest at height h above the ground. Use g for the magnitude of the acceleration of gravity. Assume that the string does not slip on the cylinder. Let v_vec represent the instantaneous velocity of the center of mass of the cylinder, and let omega_vec represent the instantaneous angular velocity of the cylinder about its center of mass. Note that there are no horizontal forces present, so for this problem vector{v} = -v (hat{y}) and vector{omega}= -omega * hat{z}. 1)Using Newton's 2nd law, complete the equation of motion in the vertical direction that describes the translational motion of the cylinder. Express your answer in terms of the tension in the vertical section of string, , and ; a positive answer indicates an upward acceleration. =? 2)Using the equation of rotational motion and the definition of torque , complete the equation of rotational motion of the cylinder about its center of mass. Your answer should include the tension in the vertical section of string and the radius . A positive answer indicates a counterclockwise torque about the center of mass (in the direction). =? 3)In other parts of this problem expressions have been found for the vertical acceleration of the cylinder and the angular acceleration of the cylinder in the direction; both expressions include an unknown variable, namely, the tension in the vertical section of string The string constrains the rotational and vertical motions, providing a third equation relating and . Solve these three equations to find the vertical acceleration, , of the center of mass of the cylinder. Express in terms of , , , and ; a positive answer indicates upward acceleration. =?Explanation / Answer
ma_y = T - mg we need to find an expression for T in terms of g, m, r, and I. We can get it from I*alpha = -Tr Tr = -I*alpha T = -I*alpha/r Omega = v/r you get: alpha = a_y/r T = -I*alpha/r T = -I*(a_y/r)/r = -I*a_y/r^2 ma_y = T - mg ma_y = -I*a_y/r^2 - mg ma_y + I*a_y/r^2 = -mg Distributive law a_y(m + I/r^2) = -mg a_y = -mg / (m + I/r^2)
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